Problem: Find all values of $x$ that satisfy \[\frac{x^2}{x+1} \ge \frac{2}{x-1} + \frac{5}{4}.\]
Explanation: Combining the terms on the right-hand side, we have \[\frac{x^2}{x+1} \ge \frac{5x+3}{4(x-1)}.\]Then, moving all the terms to the left-hand side and combining denominators again, we get \[\begin{aligned} \frac{x^2}{x+1} - \frac{5x+3}{4(x-1)} &\ge 0 \\ \frac{4x^2(x-1)-(x+1)(5x+3)}{(x+1)(x-1)} &\ge 0 \\ \frac{4x^3-9x^2-8x-3}{(x+1)(x-1)} &\ge 0. \end{aligned}\]We try to factor the numerator. Using the rational root theorem to test for rational roots, we see that $x=3$ is a root of $4x^3-9x^2-8x-3.$ Then, doing the polynomial division gives \[4x^3-9x^2-8x-3 = (x-3)(4x^2+3x+1),\]so we have \[\frac{(x-3)(4x^2+3x+1)}{(x+1)(x-1)} \ge 0.\]Since $4x^2+3x+1$ has a positive $x^2$ coefficient, and its discriminant is $3^2 - 4 \cdot 4= -7,$ which is negative, it follows that $4x^2 + 3x + 1 > 0$ for all $x.$ Thus, the above inequality is equivalent to \[f(x) = \frac{x-3}{(x+1)(x-1)} \ge 0.\]We make a sign table for $f(x)$: \begin{tabular}{c|ccc|c} &$x-3$ &$x+1$ &$x-1$ &$f(x)$ \\ \hline$x<-1$ &$-$&$-$&$-$&$-$\\ [.1cm]$-1<x<1$ &$-$&$+$&$-$&$+$\\ [.1cm]$1<x<3$ &$-$&$+$&$+$&$-$\\ [.1cm]$x>3$ &$+$&$+$&$+$&$+$\\ [.1cm]\end{tabular}We see that $f(x) > 0$ when $-1 < x < 1$ or $x > 3.$ Since the inequality is nonstrict, we also include the values of $x$ such that $f(x) = 0,$ that is, only $x = 3.$ Therefore, the solution to the inequality is \[x \in \boxed{(-1, 1) \cup [3, \infty)}.\]